A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is
1 answer:
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and
The sum is represented as
For the the values given to us the sum is calculated as
Now the since the uncertainity inthe sum is
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals meters
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Answer:
Fnet = 0
Explanation:
- Since the block slides across the floor at constant speed, this means that it's not accelerated.
- According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
- This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:
- In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:
⇒ 169 N + Fn = Fg = 216 N (3)
- This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
- Fn = 216 N - 169 N = 47 N (4)